4.5 Rectifier Circuitsjstiles/312/handouts/section_3_5...3/1/2012 section_3_5_Rectifier_Circuits.doc 1/3 Jim Stiles The Univ. of Kansas Dept. of EECS 4.5 Rectifier Circuits Reading

3/1/2012 section_3_5_Rectifier_Circuits.doc 1/3

Jim Stiles The Univ. of Kansas Dept. of EECS

4.5 Rectifier Circuits

Reading Assignment: pp. 194-200

A. Junction Diode 2-Port Networks

Consider when junction diodes appear in a 2-port network (i.e.,

a circuit with an input and an output).

We can characterize a 2-port network with its transfer

function.

HO: THE TRANSFER FUNCTION OF DIODE CIRCUITS

Finding this transfer function is similar to our previous diode

circuit analysis—but with a few very important differences!

HO: STEPS FOR FINDING A JUNCTION DIODE CIRCUIT

TRANSFER FUNCTION

EXAMPLE: DIODE CIRCUIT TRANSFER FUNCTION

+

- Ov t ( )Iv t Junction

Diode

Circuit



The input voltage is almost always a function of time, which

means the output voltage is as well.

HO: TIME-DOMAIN ANALYSIS OF DIODE CIRCUITS

B. Diode Rectifiers

Many important diode circuits appear in a standard AC to DC

power supply!

HO: POWER SUPPLIES

The signal rectifier is an important component of a power

supply—it’s what creates the DC component.

HO: SIGNAL RECTIFICATION

One standard rectifier design is called the full-wave

rectifier.

HO: THE FULL-WAVE RECTIFIER

But the bridge rectifier also provides full-wave rectification.

HO: THE BRIDGE RECTIFIER

The bridge rectifier is more complex, but results in a lower

Peak Inverse Voltage.



HO: PEAK INVERSE VOLTAGE

Make sure you can determine the PIV of a diode circuit. It

depends on both the specific circuit, and the specific input

voltage!

EXAMPLE: PEAK INVERSE VOLTAGE

2/22/2012 The Transfer Function of Diode Circuits present 1/7


The Transfer Function of Diode Circuits

For many junction diode circuits, we find that one of the voltage sources is in fact unknown! This unknown voltage is typically some input signal of the form vI, which results in an output voltage vO.

Q: How the heck do you expect us to determine vO if we have no idea what vI is??

Ov

Iv +

-

Junction Diode Circuit



The (large-signal) transfer function A: We of course cannot determine an explicit value or expression for vO, since it depends on the input vI. Instead, we will attempt to explicitly determine this dependence of vO on vI! In other words, we seek to find an expression for vO in terms of vI. Mathematically speaking, our goal is to determine the function:

O Iv f v

We refer to this as the (large-signal) circuit transfer function.

Ov

Iv +

-




Plotting the transfer function Note that we can plot a circuit transfer function on a 2-dimensional plane, just as if the function related values x and y (e.g. y f x ). For example, say our circuit transfer function is:

3 2

O I

I

v f vv

Note this is simply the equation of a line (e.g., 3 2y x ), with slope m =3 and y-intercept b =2.

vO

vI 2

3

If v



Actually, a rare moment when I’m not being annoying and pretentions

A: Actually no! Although a function is a mathematical equation, there are in fact scads of equations relating vO and vI that are not functions!

The set of all possible functions y f x are a subset of the set of all possible equations relating y and x.

A function O Iv f v is a mathematical expression such that for any value of vI (i.e., Iv ), there is one, but only one, value vO.

Q: A “function” eh? Isn’t a “function” just your annoyingly pretentious way of saying we need to find some mathematic equation relating vO and vI?



The transfer function must be a function! Note this definition of a function is consistent with our physical understanding of circuits—we can place any voltage on the input that we want (i.e., Iv ), and the result will be one specific voltage value vO on the output. Therefore, examples of valid circuit transfer functions include:

vO

vI

vO

vI



These are functions—NOT! Conversely, the transfer “functions” below are invalid—they cannot represent the behavior of circuits, since they are not functions!

vO

vI

vO

vI



The transfer function must be continuous Moreover, we find that circuit transfer functions must be continuous. That is, vO cannot “instantaneously change” from one value to another as we increase (or decrease) the value vI.

vO

vI

vO

vI

A Discontinuous Function

(Invalid circuit transfer function)

A Continuous Function

(Valid circuit transfer function)

Remember, the transfer function of every junction diode circuit must be a continuous function. If it is not, you’ve done something wrong!

2/22/2012 Steps for Finding a Junction Diode Circuit Transfer Function 1/10


Steps for Finding a Junction Diode Circuit

Transfer Function Determining the transfer function of a junction diode circuit is in many ways very similar to the analysis steps we followed when analyzing previous junction diode circuits (i.e., circuits where all sources were explicitly known). However, there are also some important differences that we must understand completely if we wish to successfully determine the correct transfer function! Step1: Replace all junction diodes with an appropriate junction diode model.

Just like before! We will now have an IDEAL diode circuit.

Step 2: ASSUME some mode for all ideal diodes.

Just like before! An IDEAL diode can be either forward or reverse biased.



Step 3: ENFORCE the bias assumption.

Just like before! ENFORCE the bias assumption by replacing the ideal diode with short circuit or open circuit.

Step 4: ANALYZE the remaining circuit.

Sort of, kind of, like before! 1. If we assumed an IDEAL diode was forward biased, we must determine i

Di —just like before! However, instead of finding the numeric value of i

Di , we determine i

Di as a function of the unknown source (e.g., i

D Ii f v ). 2. Or, if we assumed an IDEAL diode was reversed biased, we must determine i

Dv —just like before! However, instead of finding the numeric value of i

Dv , we determine i

Dv as a function of the unknown source (e.g., i

IDv f v ). 3. Finally, we must determine all the other voltages and/or currents we are interested in (e.g., vO)—just like before!



However, instead of finding its numeric value, we determine it as a function of the unknown source (e.g.,

O Iv f v ). Step 5: Determine WHEN the assumption is valid. A: Actually, no! This step is very different from what we did before! We cannot determine IF 0i

Di (forward bias assumption), or IF 0i

Dv (reverse bias assumption), since we cannot say for certain what the value of i

Di or iDv is!

Recall that i

Di and iDv are functions of the unknown voltage

source (e.g., iIDi f v and i

IDv f v ). Thus, the values of i

Di or iDv are dependent on the unknown

source (vI, say). For some values of vI, we will find that 0i

Di or 0iDv , and so

our assumption (and thus our solution for O Iv f v ) will be! correct

Q: OK, we get the picture. Now we have to CHECK to see if our IDEAL diode assumption was correct, right?



However, for other values of vI, we will find that 0iDi or

0iDv , and so our assumption (and thus our solution for

O Iv f v ) will be incorrect! A: Instead of determining IF our assumption is correct, we must determine WHEN our assumption is correct! In other words, we must determine for what values of vI is

0iDi (forward bias), or for what values of vI is 0i

Dv (reverse bias). We can do this since we earlier (in step 4) determined the function i

D Ii f v or the function iD Iv f v .

Perhaps this step is best explained by an example. Let’s say we assumed that our ideal diode was forward biased and, say we determined (in step 4) that vO is related to vI as:

2 3

O I

I

v f vv

Q: Yikes! What do we do? How can we determine the circuit transfer function if we can’t determine IF our ideal diode assumption is correct??



Likewise, say that we determined (in step 4) that our ideal diode current is related to vI as:

5

4

iID

I

i f vv

Thus, in order for our forward bias assumption to be correct, the function i

IDi f v must be greater than zero:

00

5 04

iD

I

I

if vv

We can now “solve” this inequality for vI:

5 04

5 05

I

I

I

v

vv

A: NO! Recall that vI can be any value.

Q: What does this mean? Does it mean that vI is some value greater than 5.0V ??



What the inequality above means is that 0iDi (i.e., the ideal

diode is forward biased) WHEN 5 0iDv . .

Thus, we know 2 3O Iv v is valid WHEN the ideal diode is forward biased, and the ideal diode is forward biased WHEN (for this example) 5 0i

Dv . . As a result, we can mathematically state that:

2 3 when 5 0 V O I Iv v v . Conversely, this means that if 5 0Iv . V, the ideal diode will be reverse biased—our forward bias assumption would not be valid, and thus our expression 2 3O Iv v is not correct ( 2 3 for 5 0VO I Iv v v . )! A: Time to move to the last step ! Step 6: Change assumption and repeat steps 2 through 5 ! For our example, we would change our bias assumption and now ASSUME reverse bias.

Q: So how do we determine vO for values of vI < 5.0 V ?



We then ENFORCE 0iDi , and then ANALYZE the circuit to

find both iIDv f v and a new expression O Iv f v (it will

no longer be 2 3O Iv v !). We then determine WHEN our reverse bias assumption is valid, by solving the inequality 0i

IDv f v for vI. For the example used here, we would find that the IDEAL diode is reverse biased WHEN 5 0 VIv . . For junction diode circuits with multiple diodes, we may have to repeat this entire process multiple times, until all possible bias conditions are analyzed. If we have done our analysis properly, the result will be a valid continuous function! That is, we will have an expression (but only one expression) relating vO to all possible values of vI. This transfer function will typically be piecewise linear. An example of a piece-wise linear transfer function is:



Just to make sure that we understand what a function is, note that the following expression is not a function:

vO

vI

12

5

2 3Iv

12 Iv

vO

vI

12

3

2 3Iv

12 Iv

7

2 3 5 0

12 5 0

I I

O

I I

v for v .v

v for v .

2 3 7 0

12 3 0

I I

O

I I

v for v .v

v for v .



Nor is this expression a function:

Finally, note that the following expression is a function, but it is not continuous:

vO

vI

12

3

2 3Iv

12 Iv

7

vO

vI

5

5

2 3Iv 5 Iv

2 3 3 0

12 7 0

I I

O

I I

v for v .v

v for v .

2 3 5 0

5 5 0

I I

O

I I

v for v .v

v for v .



Make sure that the piece-wise transfer function that you determine is in fact a function, and is continuous!

2/27/2012 Example Diode Circuit Transfer Function 1/7


Example: Diode Circuit

Transfer Function

Consider the following circuit, called a half-wave rectifier:

Let’s use the CVD model to determine the output voltage vO in

terms of the input voltage vI.

Ov

SV

Oi Ii

Iv

R

Di

+ -Dv

Ov

SV

Oi Ii

Iv

R

iDi

+ -iDv

0 7 .



In other words, let’s determine the diode circuit transfer

function O Iv f v !

ASSUME the ideal diode is forward biased, ENFORCE 0iDv .

From KVL, we find that:

0 0 7 0 7I O O Iv . v v v .

This result is of course true if our original assumption is correct—

it is valid if the ideal diode is forward biased (i.e., 0iDi )!

From KCL and Ohm’s Law, we find that:

0 7i O ID O

v v .i i

R R

Q: I’m so confused! Is this current greater than zero or less than zero? Is our assumption correct? How can we tell?

Ov

SV

Oi Ii

Iv

R

iDi

0+ -

0 7 .



A: The ideal diode current is dependent on the value of source

voltage Iv . As such, we cannot determine if our assumption is

correct, we instead must find out when our assumption is correct!

In other words, we know that the forward bias assumption is

correct when 0iDi . We can rearrage our diode current

expression to determine for what values of source voltage vS this

is true:

0

0 70

0 7 0

0 7

iD

I

I

I

i

v .

Rv .

v .

So, we have found that when the source voltage vI is greater than

0.7 V, the output voltage vO is:

0 7O Iv v .

Q: OK, I’ve got this result written down. However, I still don’t know what the output voltage vO is when the source voltage vI is less than 0.7V!?!



Now we change our assumption and ASSSUME the ideal diode in

the CVD model is reverse biased, an assumption ENFORCEd with

the condition that 0iDi (i.e., an open circuit).

From KCL:

0iO Di i

From Ohm’s Law, we find that the output voltage is:

0

0 V !!!

O Ov R i

R

Q: Fascinating! The output voltage is zero when the ideal diode is reverse biased.

But, precisely when is the ideal diode reverse biased? For what values of vI does this occur ?

Ov

SV

Oi Ii

Iv

R

0

+ -iDv

0 7 .



A: To answer these questions, we must determine the ideal diode

voltage in terms of vS (i.e., iD Iv f v ).

From KVL:

0 7iI D Ov v . v

Therefore:

0 7

0 7 0 0

0 7

iD I O

I

I

v v . v

v . .

v .

Thus, the ideal diode is in reverse bias when:

0

0 7 0

iD

I

v

v .

Solving for vI, we find:

0 7 0

0 7 VI

I

v .

v .

Ov

SV

Oi Ii

Iv

R

0

+ -iDv

0 7 .



In other words, we have determined that the ideal diode will be

reverse biased when 0 7 VIv . , and that the output voltage will

be 0Ov .

A: That’s right! The transfer function for this circuit is

therefore:

0 7 0 7

0 0 7

S S

O

S

v . for v .

v

for v .

Q: So, I see we have found that:

0 7 0 7 VO I Iv v . when v .

and,

0 0 0 7 VO Iv . when v .

It appears we have a valid, continuous, function!

vO

vS

1

0.7 V



Like E = mc2, the circuit in this example may seem trivial, but it is actually very important! This circuit is called a half-wave rectifier, and provides signal rectification. Rectifiers are an essential part of every AC to DC power supply!

2/27/2012 Time-domain analysis of diode circuits present 1/11


Time-domain Analysis of Diode Circuits

Let’s consider what happens when the input to a junction diode circuit varies with time! If we know the large signal transfer function of diode circuit:

( )O Iv f v= Then the time-varying output (assuming the input is not changing too fast!) is expressed as:

( ) ( )( )O Iv f vt t=

Ov t

Iv t +

-


The output will likewise vary with time!



For example… For example, say our circuit transfer function is:

2 2 5

O I

I

v f vv .

And, the input signal varies with time as:

( ) [ ]0.5cosIv t π t=

t

vO

vI 2.5

If v



Make this make sense to you

The output signal is thus: ( ) ( )

[ ]( )[ ]

2 12 10.5coscos 1

IOπ t

vv

π

t t

t

= +

= +

= +

t ( ) [ ]0.5cosIv t π t=

( ) [ ]cos 1Ov t π t= +

Ov t

Iv t +

- ( ) 2 1I If v v= +



A trickier example… Or, consider a diode circuit with this transfer function:

6.0 6.0

0 6.0

10.0 0

10.0 10.0

I

I I

O

I I

I

for v

v for vv

v for v

for v

+6

+6

vO

vI

+10

-10

-1

+1



What is the output? Say that this time-varying signal appears at its input:

+4

+8

( )Iv t

t

+12

-4

-8

-12

1 2

3 4

Ov t

Iv t +

-

vO

vI

+10

-10

-1 +1

+6

+6 What then is this output voltage?



The answer is simple—if you think Note that for time 0 1t , the input voltage is 4.0 V Since 4.0Iv volts is greater than zero, but less than 6.0 V, (0 6.0Iv ) the output voltage during this time period is 4.0O Iv v V :

0 6.

6.0 6.0

10.0 0

10.0 10.0

04.0

I

I

O

I

I

I

I

for v

vv for v

for v

v for v

vO

vI

+4

+10

-10

-1

+4 +6

+6

+4

+8

vI

t

+12

-4

-8

-12

1 2

3 4



What is the output now? Now for time 1 2t , the input voltage is 8.0 V Since 8.0Iv volts is greater than 6.0 V, ( 6.0Iv ) the output voltage during this time period is 6.0Ov V :

6.0 6.0

6.00 5.0

10.0 0

10.0 10.0

I I

O

I I

I

I

v for vv

v for

for

v

for v

v

+4

+8

vI

t

+12

-4

-8

-12

1 2

3 4

vO

vI

+10

-10

-1

+1

+8 +6

+6



The hardest one yet! Now for time 2 3t , the input voltage is -4.0 V. Since 4.0Iv volts is less than 0 V but greater than -10.0 V, ( 10 0Iv ) the output voltage during this time period is 4.0O Iv v V :

6.0 6.0

0 6.0

10

4.010.0 0

.0 10.0

I I

I

I I

O

I

v for

for v

v for vv

for v

v

+4

+8

vI

t

+12

-4

-8

-12

1 2

3 4

vO

vI

+10

-10

-1

+1

-4

+4

+6

+6



One last bit of scholarly effort Now for time 3 4t , the input voltage is -12.0 V Since 12.0Iv volts is less than -10.0 V, ( 10.0Iv ) the output voltage during this time period is 10.0Ov V :

6.0 6.0

10.0

10

0 6.0

10.0 0

.0 10.0

I

I I

O

I

I I

for v

v for vv

v fo

o

v

f v

r

r

+4

+8

vI

t

+12

-4

-8

-12

1 2

3 4

vO

vI +6

+6

+10

-10

-1

+1

-12



Put the pieces together; our output is revealed

Thus, we find that the output signal is:

That is, the output when this is the input signal.

+4

+8

( )Iv t

t

+12

-4

-8

-12

1 2

3 4

+4 +6

( )Ov t

t

+10

-4

-8

1 2 3 4



Make this make sense to you

Ov t

Iv t +-

vO

vI

+10

-10

-1 +1

+6

+6

( )Iv t

t

( )Ov t

t

2/27/2012 Power Supplies 312 present 1/22


Power Supplies Most modern electronic circuits and devices require one or more relatively low DC voltages (e.g., 5.0 V ) for biasing and for supplying power to the circuit. Big Problem Our electrical power distribution system provides a high-voltage AC output—a 120 Vrms, 60Hz sinewave!

t 5 V

VL

t

170

-170

0

( )acv t



The power supply Thus, a major component in electronic systems (e.g., computers, televisions, etc.) is the power supply. The purpose of the power supply is simply to convert high voltage AC into one or more low DC voltages.

rectifier filter voltage regulator

Step-down Transformer A Typical Power Supply



The functional schematic of a power supply A typical power supply consists of four majors sections:

1. Step-Down Transformer

2. Signal Rectifier

3. Filter

4. Voltage Regulator Let’s look at each of these devices individually!

1.





1. The Step-Down Transformer

A voltage 120 Vrms is simply too big!

The first thing to be accomplished it to “step-down” the AC to a more manageable (e.g., safe) level. We do this with a step down transformer.

1N 2N LR +

-

( )1i t

μ

( )1v t ( )2v t+

-

( )2i t



You remember—don’t you? You will recall that the important design parameters of a transformer are the number of “turns” on each side of the transformer. The integer value N1 represents the number of turns on the primary side, whereas N2 represents the number of turns on the secondary side. The voltage on the secondary side is related to the voltage on the primary side as:

( ) ( )22 1

1

Nv t v tN

=

Thus, if we wish to step-down the voltage (i.e., make 2 1v t v t ), we need to make the number of primary turns larger than the number of secondary turns (i.e., 1 2N N ). Typically, a step-down transformer will lower the AC voltage to around 30

Vrms.



And Michael Faraday never even attended college—so now what’s your excuse?

Remember, a transformer is a fundamental application of Faraday’s Law—one of Maxwell’s equations!

( ) ( )1 1

,C S

r d r t dst¶

⋅ = - ⋅¶ò òòE B

( ),r tB

1N 2N LR

( )1i t

( )1v t ( )2v t+

-

( )2i t

+-



Pure AC; no DC Recall that the output of the step-down transformer is a sinusoid—a “pure” AC signal. In other words, the output has no DC component. But this is a problem, as we need a DC signal at the power supply output!

t

170

-170

0

( )acv t

t 5 V

VL



2.The Signal Rectifier This is where the signal rectifier comes in—its job is to take an AC signal, and produce a signal with a DC component. Q: So, the output of a rectifier is a DC signal? A: NO! I didn’t say that! The output of a rectifier has a DC component. However, it likewise has an AC component—the signal output is still time varying!

rectifier

Step-down Transformer

( )rctOv t+

-

( )acv t+

-



AC/DC Q: Huh? A: Most signals are neither “purely” AC (a time-varying signal with a time-averaged value of zero), or “purely” DC (a constant with respect to time). Most signals are a combination of AC and DC—they have both an AC component and a DC component. I.E., they can be expressed as the sum of a DC and AC signal:

DC ac tVv t v The DC component of a signal v t is simply its time-averaged value:

0

1DC

T

vV t dtT

and the AC component of a signal v t is simply the signal v t with its DC component removed:

Dac Cvv tt V



Most time-varying signals are not AC signals The important fact about an AC component is that it has a time-averaged value of zero!

0 0

0 0

0

1 1

1 1

1

10

0

T T

T T

T

DC

DC

DC DC

DC DC

DC DC

ac dt dtT T

dt dt

V

V

V V

V V

V V

v

T T

dt

v t t

T

v t

TT

Pay attention: an AC signal is defined as a time-varying signal whose average value is zero!



Please tell me this is obvious For example, a signal might have the form: ( ) 0.66 0.001cost ωtv = +

It is hopefully evident that this signal has a

DC component:

and an AC component:

0.0

0.66

v

t

0

0

0 0

1

1 0.66 0.001cos

0.66 0.001 cos

0. 066 60 .6

D

T

T

T T

C dtT

dtωtT

dt ωtT T

V v t

dt

( ) ( )

( ) 0.660.66 0.0010.001c

cosos

a DCcω

tt

v v t

ω

V

t

= -

= -+=



A time-varying signal with an DC component

Since the input to the rectifier is a 60Hz “sine wave”, it has no DC component—the time averaged value of a sine function is zero. Thus, the input is a “pure” AC signal. The output of a rectifier is likewise time-varying—it has an AC component. However, the time-averaged value of this output is non-zero—the output also has a DC component!

t

A

rctIv t

-A

0

rctOv t v



The Rectifier creates a DC component; but the AC component is still there

Thus, a rectifier creates a DC component on the output, a component that does not exist on the input!

2DC

AV

t

A ( ) ( )D

rctO acCVv v tt = + v

vac(t) 0



3. The Filter So, the rectifier has added a DC component to the signal—but the output of the rectifier is not DC, it has an AC component as well. The job of any filter is to remove unwanted components, while allowing the desired components to pass through. For this electrical filter, the unwanted component is the AC signal, and the desired component is the DC signal!

We do this with what is essentially a low pass filter.

rectifier filter




The DC component passes through the filter unchanged

Ideally, the low-pass filter would eliminate the AC component, leaving only a constant, DC signal. However, a low-pass filter will typically just attenuate the AC signal, leaving a small AC component at the filter output. For example, the ouput of the rectifier is now the input to filter: The DC component passes through the low-pass filter unscathed—it’s the

same at the output as the input.

2DC

AV

t

A ( ) ( )D

fltI acCVv v tt = + v

vac(t) 0



Ripple voltage! But; the AC component is greatly attenuated. Thus, the output is almost a DC signal—it is a DC signal with a small AC “ripple voltage”. Q: Yikes! Our load (e.g., computer, HDTV, DVR, DVD, etc.) will require a better regulated voltage than that! Won’t the “ripple voltage” cause all sorts of problems—if it is used to supply power to the load! A: Yes, and it’s even worse than that! You see, the 120 Vrms , 60 Hz signal from our wall socket may not be exactly

that!

2DC

AV

t

A ( ) ( )D

fltO acCVv v tt = + v

vac(t) 0



It will be 60 Hz; it might not be 120 Vrms Q: You mean it might not be exactly 60 Hz? A: Actually, a 60 Hz sinusoid is the one thing we can count on—the AC power signal will have a frequency of exactly, precisely 60Hz.

However, the magnitude of this AC power signal (nominally 120 Vrms ) is a bit more problematic. Power lines, transformers—all the stuff that the AC power must pass through—exhibit resistance.

The more current passing through the power system, the more this resistance results in a voltage drop (it’s just Ohm’s Law at work!). As a result, the AC voltage at your wall plug will be only approximately 120

Vrms.



When all those air conditioners are “on” Say after a brutally hot July day, it’s still 93 degrees F at 8:30 p.m. The whole town is at home; lights on, watching TV, playing Wii. And, most importantly, running their air conditioners! All this stuff requires energy—for this case, energy delivered at a particularly high rate.

Thus, the AC current flowing into all these cool and content homes is particularly large, meaning that the “Ohmic Loss”—the voltage drop that occurs between the power plant to your home—is likewise particularly large. Therefore, the AC voltage at your wall socket on this sweltering day may be significantly less than 120 Vrms !



It’s all proportional to the AC power voltage Q: Is this a problem? A: Absolutely! If the voltage of the AC power lessens, a proportionate reduction will occur at the output of the step-down transformer. Therefore the DC component at the output of the rectifier (and thus at the output of the filter) will also reduce proportionately! Q: So in addition to the ripple voltage, the DC component of the signal can drift up or down—isn’t this DC supply voltage horribly regulated? A: That’s exactly correct—which is why we need, as the last component of the power supply, a voltage regulator!



4. The Voltage Regulator This regulator of course provides at its output a regulated DC voltage—a voltage that stays a rock-solid constant, regardless of how many air-conditioners are running (i.e., great line regulation). And, regardless of the current drawn by the load (great load regulation). This voltage regulator could of course either be a linear or switching regulator.





We often need several regulators Moreover, we find that the power supplies of most complex loads (e.g., computers, TVs, etc.) employ components that require several different DC source voltages. As a result, power supplies often use many regulators, each with a different DC voltage :

5.0V

12.0V

3.5V

rectifier filter

voltage regulator

#1

voltage regulator

#2

voltage regulator

#3



This circuit will be on the final The power supply: make sure you know why it’s needed; what it does; and how it does it!

Copyright © 1994-2012 Samuel M. Goldwasser --- All Rights Reserved ---

http://www.repairfaq.org/sam/samschem.htm

2/27/2012 Signal Rectification present 1/15


Signal Rectification An important application of junction diodes is signal rectification. There are two types of signal rectifiers, half-wave and full-wave. Let’s first consider the ideal half-wave rectifier. It is a circuit with the transfer function O Sv f v :

vO

vI

1 0 0

0

I

O

I I

for vv

v for v

Ov t

Iv t Ideal

½ Wave Rectifier

+-



It’s the same—or zero

Pretty simple! When the input is negative, the output is zero, whereas when the input is positive, the output is the same as the input.

A: To see why a half-wave rectifier is useful, consider the typical case where the input source voltage is a sinusoidal signal with frequency ω and peak magnitude A:

Q: Pretty pointless I’d say.

This appears to be your most useless circuit yet.

t

A

Iv t

-A

0

sinIv A ωtt



Half a sine wave

Think about what the output of the half-wave rectifier would be! Remember the rule: when vI (t) is negative, the output is zero, when vI (t) is positive, the output is equal to the input.

The output of the half-wave rectifier for this example is therefore:

t

A

vI(t) -A

0

vO(t)

v

Q: A half a sine wave? What kind of lame signal is that?



The input is a pure AC signal… A: Although it may appear that our rectifier had little useful effect on the input signal vI(t), in fact the difference between input vI(t) and output vO(t) is both important and profound. To see how, consider first the DC component (i.e. the time-averaged value) of the input sine wave:

0

0

1

1 sin 0

T

I I

T

V v dttT

A ωt dtT

Thus, (as you probably already knew) the DC component of a sine wave is zero—a sine wave is an AC signal!



…but the output has a DC component as well Now, contrast this with the output vO(t) of our half-wave rectifier.

The DC component of the output is:

Unlike the input, the output has a non-zero (positive) DC component ( DCV A π )!

Q: I see. A non-zero DC component eh? So refresh my memory, why is that important?

0

2

02

1

1 1sin 0

T

DC O

T T

T

V v dttT

A ωt dt dt AT T π

t

A

-A

DCAV π

vO(t) v



A DC component is required A: Recall that the power distribution system we use is an AC system. The source voltage vI(t) that we get when we plug our “power cord” into the wall socket is a 60 Hz sinewave—a source with a zero DC component! The problem with this is that most electronic devices and systems, such as TVs, stereos, computers, etc., require a DC voltage(s) to operate! A: That’s why the half-wave rectifier is so important! It takes an AC source with no DC component and creates a signal with both a AC and DC component.

Q: But, how can we create a DC supply voltage if our power source vI(t) has no DC component??



Remember this?

We can then pass the output of a half-wave rectifier through a low-pass filter, which suppresses the AC component but lets the DC value ( OV A ) pass through. We then regulate this output and form a useful DC voltage source—one suitable for powering our electronic systems!





An ideal rectifier requires an ideal diode

A: An ideal half-wave rectifier can be “built” if we use an ideal diode.

If we follow the transfer function analysis steps we studied earlier, then we will find that this circuit is indeed an ideal half-wave rectifier!

Q: OK, now I see why the ideal half-wave rectifier might be useful. But, is there any way to actually build this magical device?

vO

vS

1

Ov t

Sv +- Iv t

R

iDi

+ -iDv

0 0

0

I

O

S I

for vv

v for v



But a junction diode works as well Of course, since ideal diodes do not exist, we must use a junction diode instead: A: Yes! It was an example where we determined the junction diode circuit transfer function.

Q: This circuit looks so familiar! Haven’t we studied it before?

Ov t

Sv +- Iv t

R

Di

+ -Dv



It’s a little different from ideal Recall that the result was: Note that this result is slightly different from that of the ideal half-wave rectifier! The 0.7 V drop across the junction diode causes a horizontal “shift” of the transfer function from the ideal case. A: Hardly! Although the transfer function is not quite ideal, it works well enough to achieve the goal of signal rectification—it takes an input with no DC component and creates an output with a significant DC component!

vO

vI

1

0.7 V

Q: So then this junction diode circuit is worthless?

0 7 0 7

0 0 7

I I

O

I

v . for v .v

for v .



Let’s determine the output Note what the transfer function “rule” is now:

1. When the input is greater than 0.7 V, the output voltage is equal to the input voltage minus 0.7 V.

2. When the input is less than 0.7 V, the output voltage is zero.

So, let’s consider again the case where the source voltage is sinusoidal (just like the source from a “wall socket”!):

sin 120Iv A π tt

t

A

vs(t)

-A

0.7



Close to ideal—and still plenty of DC The output of our junction diode half-wave rectifier would therefore be: Although the output is shifted downward by 0.7 V (note in the plot above this is exaggerated, typically A >> 0.7 V), it should be apparent that the output signal vO(t), unlike the input signal vI(t), has a non-zero (positive) DC component. Because of the 0.7 V shift, this DC component is slightly smaller than the ideal case. In fact, we find that if A>>0.7, this DC component is approximately:

0.35DCAV Vπ

In other words, just 350 mV less than ideal!

t

A

vI(t) -A

vO(t)

v

0.7



The ideal full-wave rectifier

A: Almost forgot! Let’s examine the transfer function of an ideal full-wave rectifier:

Q: Way back on the first page you said that there were two types of rectifiers. I now understand half-wave rectification, but what about these so-called full-wave rectifiers?

0

0

I I

O

I I

v for vv

v for v

vO

vI

1 -1



The DC component is twice as big If the ideal half-wave rectifier makes negative inputs zero, the ideal full-wave rectifier makes negative inputs—positive! For example, if we again consider our sinusoidal input, we find that the output will be: The result is that the output signal will have a DC component twice that of the ideal half-wave rectifier!

0

2

02

1

1 1sin si 2n

T

DC O

T T

T

V v dttT

A ωt dt A ωt dtT

AπT

t

A

vI(t) -A

0

vO(t) v



We can build a nearly ideal full-wave rectifier with junction diodes

A: Although we cannot build an ideal full-wave rectifier with junction diodes, we can build full-wave rectifiers that are very close to ideal with junction diodes!

Q: Wow! Full-wave rectification appears to be twice as good as half-wave. Can we build an ideal full-wave rectifier with junction diodes?

t

A

-A

2DC

AV π

vO(t) v

2/29/2012 The Full Wave Rectifier present 1/23


The Full-Wave Rectifier Consider the following junction diode circuit:

Note that we are using a transformer in this circuit. The job of this transformer is to step-down the large voltage on our power line (120 Vrms) to some smaller magnitude (typically 20-70 Vrms).

Power Line

D1

D2

( )Ov t+

− R( )Iv t

+

−

( )Iv t

+

−

( )ACv t

+

−



This point is very important! Note the secondary winding has a center tap that is grounded. Thus, the secondary voltage is distributed symmetrically on either side of this center tap. For example, if vI = 10 V, the anode of D1 will be 10 V above ground potential. While, the anode of D2 will be 10 V below ground potential (i.e., -10V):

R

Power Line

D1

D2

( )Ov t+

− R10 0

+

−

.

10 0

+

−

.

10 0+ .

10 0− .



Make this make sense Conversely, if vI =-10 V, the anode of D1 will be 10V below ground potential (i.e., -10V), while the anode of D2 will be 10V above ground potential:

Power Line

D1

D2

( )Ov t+

− R10 0

+

−

−

.

10 0

+

−

−

.

10 0− .

10 0+ .



Let’s redraw the circuit… The more important question is, what is the value of output vO? More specifically, how is vO related to the value of source vI —what is the transfer fuction ( )O Iv f v= ? To help simplify our analysis, we are going redraw this cirucuit in another way. First, we will split the secondary winding into two explicit pieces:

Power Line

D1

D2

( )Ov t+

− R( )Iv t

+

−

( )Iv t

+

−

( )ACv t

+

−



…nothing has changed—it’s the same circuit We will now ignore the primary winding of the transformer and redraw the remaining circuit as: Note that the secondary voltages at either end of this circuit are the same, but have opposite polarity.

D1 D2

( )Ov t+

− R( )Iv t

+

−

( )Iv t

−

+



Just like a teeter-totter; one side goes up, the other side goes down

As a result, if vI =10, then the anode of diode D1 will be 10V above ground, and the anode at diode D2 will be 10V below ground—just like before!

10 0+ . 10 0− . D1 D2

Ov+

− R10

+

−

10

−

+



And vice vesa And, if vI =-10, then the anode of diode D1 will be 10V below ground, and the anode at diode D2 will be 10V above ground—just like before!

Now, let’s attempt to determine the transfer function ( )O Iv f v= of this circuit!

10 0− . 10 0+ . D1 D2

Ov+

− R10

+

−

−

10

−

−

+



Yikes! Three things to find! First, we will replace the junction diodes with CVD models. Then let’s ASSUME D1 is forward biased and D2 is reverse biased, thus ENFORCE 1 0i

Dv = and 2 0iDi = .

Thus ANALYZE: Note that we need to determine 3 things:

* the ideal diode current 1

iDi ,

* the ideal diode voltage 2

iDv ,

* and the output voltage vO.

Ov+

− RIv

+

−

Iv

−

+

1

iDi

0 7 + −. 2- +i

Dv

0 7 − +. Oi



Sprinkle on some KVL pixie dust However, instead of finding numerical values for these 3 quantities, we must express them in terms of input voltage vI ! From KCL:

1 2 1 10i i i iO D D D Di i i i i= + = + =

From KVL:

10 0.7 0iDI R iv − − − =

Thus the ideal diode current is:

1

0.7IiD

viR−

=

2- +iDv

Ov+

− RIv

+

−

Iv

−

+

1

iDi

0 7 + −.

0 7 − +. Oi



It’s getting dusty in here

Likewise, from KVL:

20 0.7 0.7 0IiD Iv v v− − + + + =

Thus, the ideal diode voltage is:

2 0.7 0.7 2iD I IIv v vv= − − + − = −

And finally, from Ohm’s Law:

10.7 0.7I

Ii

O Dv i R RR

v v −= = = −

Ov+

− RIv

+

−

Iv

−

+

1

iDi

0 7 + −. 2- +i

Dv

0 7 − +. Oi



Does not mean that vI is bigger than 0.7

Thus, the output voltage is: 0.7IOv v= −

Now, we must determine when both 1 0i

Di > and 2 0iDv < .

When both these conditions are true, the output voltage will be 0.7IOv v= − . When one or both conditions 1 0i

Di > and 2 0iDv < are false, then our assuptions are

invalid, and 0.7IOv v≠ − . Using the results we just determined, we know that 1 0i

Di > when:

0.70Iv

R−

>

Solving for vI : 0.7

0

0.7 00.7 V

I

I

I

Rv

vv

−>

− >

>



Make this make sense

Likewise, we find that 2 0iDv < when:

2 0Iv− <

Solving for vI :

2 02 0

0

I

I

I

vvv

− <

>

>

Thus, our assumptions are correct when 0.0Iv > AND 0.7Iv > .

THINK about this. This is the same thing as saying our assumptions are valid when 0.7Iv > !

Thus, we have found that the following statement is true about this (but only this!) circuit:

0.7 when 0.7O I Iv v V v V= − >



We’re not done yet!

0.7 when 0.7O I Iv v V v V= − >

Note that this statement does not constitute a function (what about 0.7Iv < ?), so we must continue with our analysis!

0 7Iv+−−

. R0 7Iv

+

>

−

. 0 7Iv

−

>

+

.

0 0 7 + −. 0- +

0 7 − +. 0

0 7> . 0 7< − .



A good engineer is fretful, paranoid and fatalistic

Q: Wait! I’m concerned about something. We found that the voltage across the second ideal diode is:

2 2 IiD vv = −

From the CVD model, that means the voltage across junction diode D2 is approximately:

2 20.7 0.7 2iD D Ivv v= + = −

Since we know this is true only when:

0.7Iv V> the diode voltage 2 0.7 2 ID vv = − must be negative.



Avoid breakdown! Moreover, this negative voltage is proportional to twice the input voltage! Thus, if the input voltage is large, the voltage across this junction diode might be very, very negative. Shouldn’t I be concerned about this junction diode going into breakdown? A: You sure should! If the junction diode goes into breakdown, the transfer function will not be what we expected. You’d beter use junction diodes with sufficiently large Zener breakdown voltages!



Peak Inverse Voltage: more on this later Q: But how large is sufficiently large? A: If we know precisely the input voltage function ( )Iv t , we can find the “worst case” scenario—the most negative voltage value that occurs across this junction diode. We call the magnitude of this value the Peak Inverse Voltage (more on this later)—the ZKV of our Zener diodes had better be larger that this value! INVERSE

VOLTAGE



Back to the analysis; I’ll skip some stuff OK, back to the anaysis, say we now ASSUME that D1 is reverse biased and D2 is forward biased, so we ENFORCE 1 0i

Di = and 2 0iDv = .

Thus, we ANALYZE this circuit: Using the same proceedure as before, we find that 0.7O Iv v= − − , and both our assumptions are true when 0.7 VIv < − .

Ov+

− RIv

+

−

Iv

−

+

2

iDi

0 7 + −. 1+ -i

Dv

0 7 − +. Oi



And we’re still not finished! In other “words”:

0.7 V when 0.7 VO I Iv v v= − − < −

Note we are still not done! We do not have a complete transfer function (what happens when

0.7 V 0.7Iv V− < < ?).

0 7< − . 0 7> .

0 7Iv− −−

+. R0 7Iv < −

−

+

. 0 7Iv < −

+

−

.

00 7 + −.

0+ - 0 7 − +.

Oi



Now we’re done! Finally then, we ASSUME that both ideal diodes are reverse biased, so we ENFORCE 1 0i

Di = and 2 0iDi = .

Thus ANALYZE:

Following the same proceedures as before, we find that 0Iv = , and both assumptions are true when 0 7 0 7I. v .− < < . In other words:

0 when 0 7 0 7I Iv . v .= − < <

2- +iDv

0 7 − +.

Ov+

− RIv

+

−

Iv

−

+

0 7 + −. 1+ -i

Dv

Oi



A simple analysis Now we have a function!

0- + 0 7 − +.

0+

− R0 7 0 7Iv− < <

+

−

. . 0 7 0 7Iv− < <

+

−

. .

0 7 + −.

0+ - 0



Smells like an ideal full-wave rectifier! The transfer function of this circuit is:

0 7 0 7

0 0 7 0 7

0 7 0 7

I I

O I

I I

v . V for v . V

v V for . v . V

v . V for v . V

− >= − > >− − < −

Note how this compares to the transfer function of the ideal full-wave rectifier: Very similar!

vO

vS

1 -1

-0.7 0.7

vO

vI

1 -1 0

0

I I

O

I I

v for vv

v for v

− <= >



See? Likewise, compare the output of this junction diode full-wave rectifier: to the output of an ideal full-wave rectifier: Again we see that the junction diode full-wave rectifier output is very close to ideal.

t

A

vI(t) -A

vO(t) v

0.7

-0.7

t

A

vI(t) -A

0

vO(t) v



The DC component of the output is nearly ideal!

In fact, if A >> 0.7 V, the DC component of this junction diode full wave rectifier is approximately:

2 0.7 VDCAV

π≈ −

Just 700 mV less than the ideal full-wave rectifier DC component!

D1 D2

( )Ov t+

−

R( )Iv t

+

−

( )Iv t

−

+

2/29/2012 The Bridge Rectifier present 1/15


The Bridge Rectifier Now consider this junction diode rectifier circuit: We call this circuit the bridge rectifier.

Let’s analyze it and see what it does!

i

Iv t

+

Ov t

R

Power Line

_

i



16 possible assumptions! First, we replace the junction diodes with the CVD model: A: True! However, there are only three of these sets of assumptions are actually possible!

Q: Four gul-durn ideal diodes! That means 1 6 sets of dad-gum assumptions!

i

Iv t

+

Ov t

R

Power Line

_

i

D1

D2

D3

D4 0.7 V

0.7 V

0.7 V

0.7 V



But only 3 sets are possible Consider the current i flowing through the rectifier. This current of course can be positive, negative, or zero. It turns out that there is only one set of diode assumptions that would result in positive current i , one set of diode assumptions that would lead to negative current i , and one set that would lead to zero current i . A: Regardless of the value of source vS, the remaining 13 sets of diode assumptions simply cannot occur for this particular circuit design!

Q: But what about the remaining 1 3 sets of dog gone diode assumptions?

i

Iv t

+

Ov t

R

_

i

D1

D2

D3

D4 0.7 V

0.7 V

0.7 V

0.7 V



i > 0 Let’s look at the three possible sets of assumptions, first starting with 0i > . The rectifier current i can be positive only if these assumptions are true:

* Ideal diodes D1 and D3 are reverse biased.

* Ideal diodes D2 and D4 are forward biased.

Iv t

+

Ov t

R

Power Line

_

i > 0

0.7 V

0.7 V

0.7 V

0.7 V

1iDv

3iDv

2i

Di

4i

DiRi

i > 0



Breakdown: it appears to be less likely Analyzing this circuit, we find from KVL that the output voltage is:

1 4 VO Iv v .= −

and the forward biased ideal diode currents are from KCL and Ohm’s Law:

2 4

1 4i i ID D R

v .i i i iR−

= = = =

and, finally the reverse biased ideal diode voltages are from KVL:

iD Iv v= −

Q: Hey! I notice that the reverse bias voltage for this bridge rectifier is much less negative than that of the full-wave rectifier (i.e., 2i

D Iv v= − for the full-wave rectifier). Does that mean breakdown is less likely? A: Absolutely! This is an important feature of the bridge rectifier (more on this later!).



Way to be paranoid! Now, back to the analysis…

Thus, 0i

Di > when: 1 4

0

1 4 01 4 V

I

I

I

v .R

v .v .

−>

− >

>

and 0i

Dv < when: 00

I

I

vv− <

>

Therefore, we find that for this circuit that:

1 4 when 0 1 4O I I Iv v . V v v . V= − > >and Applying some logic, we see that this simply means

1 4 when 1 4O I Iv v . V v . V= − >



i < 0 The rectifier current i can be negative only if these assumptions are true:

* Ideal diodes D1 and D3 are forward biased.

* Ideal diodes D2 and D4 are reverse biased.

Iv t

+

Ov t

R

Power Line

_

i < 0

0.7 V

0.7 V

0.7 V

0.7 V 2iDv

3i

Di

1i

Di

Ri

i < 0

4iDv



Verify this yourself Analyzing this circuit, we find from KVL that the output voltage is:

1 4 VO Iv v .= − − while the forward biased ideal diode currents are both determined from KCL and Ohm’s Law:

1 3

1 4i i SD D R

v .i i i iR

− −− = = = =

and the reverse biased ideal diode voltages are found from KVL to be:

1 3i iD D Iv v v= = (remember this for later!)



Applying logic: it will get you far in life Thus, 0i

Di > when: 1 4

0

1 4 01 4 V

1 4 V

I

I

I

I

v .R

v .v .v .

− −>

− − >

− >

< −

and, 0i

Dv < when: 0Iv <

Therefore, we find that for this circuit that:

1 4 when 0 1 4O I I Iv v . V v v . V= − < < −and Applying some logic, we see that this simply means:

1 4 V when 1 4VO I Iv v . v .= − − < −



i = 0 The rectifier current i can be zero only if these assumptions are true:

All ideal diodes are reverse biased!

Iv t

+

Ov t

R

Power Line

_

i = 0

0.7 V

0.7 V

0.7 V

0.7 V 2iDv

Ri

i = 0

4iDv

1iDv

3iDv



Verify this yourself Analyzing this circuit, we find that the output voltage is:

0O Rv R i R i= = = while the ideal diode voltages of D2 and D4 are each:

2 4

1 42

i iID D

v .v v−= =

and the ideal diode voltages of D1 and D3 are each:

1 3

1 42

i iID D

v .v v− −= =



This logic is a bit simpler Thus, 2 0i

Dv < when: 1 4

02

1 4 01 4

I

I

I

v .

v .v .

−<

− <

<

and, 1 0i

Dv < when: 1 4

02

1 4 01 4

1 4

I

I

I

I

v .

v .v .v .

− −<

− − <

− <

> −

Therefore, we also find for this circuit that:

0 when both 1 4 and 1 4 O S Sv v . V v . V= < > −

Or, in other “words”:

0 when 1 4 1 4 O Iv . V v . V= − < <



It’s true: class of 1983!

A: We know that we have considered every possible case, because when we combine the three results we find that we have a piece-wise linear function!

Q: You know, that dang Mizzou grad said we only needed to consider these three sets of diode assumptions, yet I am still concerned about the other 13. How can we be sure that we have analyzed every possible set of valid diode assumptions?



Smells like a full-wave rectifier!

1 4 V 1 4V

0 -1.4 1 4V

1 4 V 1 4V

S S

O S

S S

v . if v .

v if v .

v . if v .

− − < −= < <

− >

Note that the bridge rectifier is a full-wave rectifier!

vO

vS

1 -1

-1.4 1.4



Close to the ideal result If the input to this rectifier is a sine wave, we find that the output is approximately that of an ideal full-wave rectifier: We see that the junction diode bridge rectifier output is very close to ideal. In fact, if A>>1.4 V, the DC component of this junction diode bridge rectifier is approximately:

2 1.4 VDCA

πV

Just 1.4 V less than the ideal full-wave rectifier DC component!

t

A

vI(t) -A

vO(t) v

1.4

-1.4

3/1/2012 Peak Inverse Voltage present.doc 1/15


Peak Inverse Voltage

A: First, a slight confession—the results we derived for the bridge and full-

wave rectifiers are not precisely correct!

Recall that we used the junction diode CVD model to determine the transfer

function of each rectifier circuit.

The problem is that the CVD model does not predict junction diode breakdown!

Q: I’m so confused! The bridge rectifier and the full-wave rectifier both provide full-wave rectification. Yet, the bridge rectifier use 4 junction diodes, whereas the full-wave rectifier only uses 2. Why would we ever want to use the bridge rectifier?



Doc, it hurts when I do this

If the input voltage vI becomes too large, the junction diodes can in fact

breakdown—but the transfer functions we derived do not reflect this fact!

A: Fortunately no.

Breakdown is an undesirable mode for circuit rectification.

Our job as engineers is to design a

rectifier that prevents it from

occuring—that why the bridge

rectifier is helpful!

Q: You mean that we must rework our analysis and find new transfer functions!?



Whew! That’s a big negative number

To see why, consider the voltage across a reversed biased junction diode in each

of our rectifier circuit designs.

Recall that the voltage across a reverse biased ideal diode in the full-wave

rectifier design was:

22i

D Iv v

so that the voltage across the junction diode is approximately (according to the

CVD model):

2 20.7 2 0.7i

D D Iv v v

Ov RIv Iv 1

iDi

0 7 .

2- +iDv

0 7 .

Oi



Like getting me for all your classes

Now, we wish to determine the worst case scenario, with

respect to negative diode voltage.

We seek minDv , the minimun (i.e., most negative) value that

the diode voltage will ever be—at least, for a given input

Iv t .

I.E.: minD Dv v t for all time t

The value minDv is a negative number!

Recall that for the junction diode to avoid breakdown, the diode voltage must be

greater than ZKV for all time t (i.e., D ZKv Vt ).



He’s not trying to fly; he’s indicating “safe”

The worst case scenario occurs at the time when the diode voltage is at its most

negative (i.e., when minD Dv vt ).

Thus, we know we are safe (no breakdown—ever!) IF:

minD ZKv V

Now, assume that the source voltage is a sine wave sinIv A ωtt .

We find that diode voltage is at it most negative (i.e., breakdown danger!) when

the source voltage is at its maximum value maxIv A .

Therefore:

2 0 7 2 0 7min maxD Iv v . A .

Of course, the largest junction diode voltage occurs when in forward bias:

0 7maxDv . V



Wow; that diode voltage goes way negative!

Plotting both input sinIv A ωtt and diode voltage 2Dv t for the full-wave

rectifier:

Note that this minimum diode voltage minDv is very negative, with an absolute

value ( 2 0 7minDv A . ) nearly twice as large as the source magnitude A.

t

maxIv A

vI(t)

vD2(t)

v

0.7

2 0 7

minDv

A .



Peak Inverse Voltage: it’s a positive value

Since minDv is negative, we take its magnitude, “converting “ it into a positive value

we call the Peak Inverse Voltage (PIV):

min

DPIV v

The PIV is a positive number!

For example, the PIV of this full-wave

rectifier, with a sinusoidal input is:

2 0 7minDPIV v A .

INVERSE

VOLTAGE



The input and the circuit—

PIV depends on both !

It is crucial that you understand that the Peak Inverse Voltage (PIV) is

dependent on two things:

1. the rectifier circuit design, and

2. the input voltage Iv t .

A: The Peak Inverse Voltage specifies the worst case scenario with respect to

negative diode voltage.

It allows us to answer one important question—will the junction diodes in our

rectifier breakdown?

Q: So, why do we need to determine PIV? I’m not sure I see what difference this value makes.



You’re safe if PIV is less than VZK

To avoid breakdown, we earlier found that minDv must be greater than ZKV :

minD ZKv V to avoid breakdown

Multiplying by -1, we equivalently state:

minD ZKv V to avoid breakdown

But since minDv is negative, the value min

Dv is positive, and so

min minD Dv PIVv

Inserting the result in the earlier inequality, we find that

breakdown is avoided if:

ZKPIV V



In summary:

If the PIV is less than the Zener breakdown voltage of your

rectifier diodes. I.E.,

if ZKPIV V ,

then we know that your junction diodes will remain in either

forward or reverse bias for all time t.

Your rectifier will operate “properly”!

However, if the PIV is greater than the Zener breakdown

voltage of your rectifier diodes. I.E.:

if ZKPIV V ,

then you know that our junction diodes will breakdown for at least

some small amount of time t.

Then the rectifier will NOT operate properly!



But what if PIV is too big?

A: We have three possible solutions:

1. Use junction diodes with larger values of VZK (but they exist!).

2. Reduce the input voltage (e.g., magnitude A), but this will decrease you DC

component DCV

3. Use the bridge rectifier design—no buts about it!

Q: So what do we do if PIV is greater than VZK ? How do we fix this problem?



The bridge rectifier to the rescue

A: To see how a bridge rectifier can be useful, let’s determine its Peak Inverse

Voltage PIV.

Q: The bridge rectifier! How does that solve our breakdown problem?

Iv t

+

Ov t

R

Power Line

_

i > 0

0.7 V

0.7 V

0.7 V

0.7 V

1iDv

3iDv

2i

Di

4i

Di

Ri

i > 0



Danger, danger!

First, we recall that the voltage across a reverse biased ideal diode was:

1iD Sv v

so that the voltage across the junction diode is approximately:

1 10.7

0.7

iD D

I

v v

v

Now, assume that the source voltage is a sine wave sinIv A ωtt .

We found that diode voltage is at it most negative (i.e., breakdown danger!) when

the source voltage is at its maximum value maxIv A .

I.E.,:

0 7 0 7min maxD Iv v . A .



Hmm… It’s negative, but not extremely so

Of course, the largest junction diode voltage occurs when in forward bias:

0 7 Vmax

Dv .

Plotting both input sinIv A ωtt and diode voltage 1Dv t for the bridge

rectifier:

Note that this minimum diode voltage is negative, with an absolute value

( 0 7minDv A . ), approximately equal to the value of the source magnitude A.

t

A vI(t)

vD1(t)

v

0.7

0 7A .



The bridge rectifier PIV is

about half the full-wave PIV

Thus, the PIV for a bridge rectifier with a sinusoidal source voltage is:

min 0.7brg DPIV Av

Note that this bridge rectifier value is approximately half the PIV we

determined for the full-wave rectifier design:

min 2 0.7fw DPIV Av

Thus, the source voltage (and the output DC component) of a bridge rectifier can

be twice that of the full-wave rectifier design.

This is why the bridge rectifier is a

very useful rectifier design!

3/1/2012 Example Peak Inverse Voltage.doc 1/11


Example: Peak

Inverse Voltage

The diode in the circuit below is IDEAL.

The input to this circuit ( Iv t ) is plotted below.

Let’s determine the Peak Inverse Voltage (PIV) of the ideal

diode in this circuit.

Carefully

consider this

scale!

+6

+12

vI (t)

t -6

-12

+18

R1 =3K R2 =1K

vI (t)

2.0 V



Q: This is so confusing! Is this the right equation:

0 7PIV A . Or, do I use this equation:

2 0 7PIV A . ?? Likewise, is A=18, or is A=12, or is it some other number ?

A: None of the above!

The result:

0 7PIV A .

is the PIV for the following specific situation—and this

situation only:

1. the diode circuit is a bridge rectifier, with

2. an input signal sinIv A ωtt

Of course, the circuit for this problem is most definately not

a bridge rectifier (it doesn’t even have an output!):



And the input signal is most definitely not a sinusiod.

Thus, 0 7PIV A . is most definitely not “the right

equation”!

Q: Oh, so then I should use the other one:

2 0 7PIV A . ?

A: Ahem; the result:

2 0 7PIV A .

is the PIV for this specific situation—and this situation only:

+6

+12

vI (t)

t -6

-12

+18

R1 =3K R2 =1K

vI (t)

2.0 V



1. the diode circuit is a full-wave rectifier, with

2. an input signal sinIv A ωtt

Of course, the circuit in this problem is not a full-wave

rectifier (it doesn’t even have an output!), and the input signal

is not a sinusiod.

Thus, you most definitely do not “use” the equation

2 0 7PIV A . !

Q: Apparently, you could “use” a book on teaching, because these are the only two PIV equations that you gave us!

A: I did not “give” you these

equations—they appeared as a result of

a careful and detailed analysis of the

specific situations we encountered.

You now have encountered (with this problem) a completely

different situation—you can find the “right equation to use”

only after you carefully, patiently, completely analyze this

new, specific situation!

To begin, you first ASSUME that the ideal diode is reverse

biased, and thus ENFORCE the condition that 0iDi :



From KCL:

2 1

1

1

0

iDi i i

i

i

From Ohm’s Law and the results above:

1 1 1 1 2 2 2 2 13 1 1v i R i v i R i i

Now from KVL:

1 2 0Iv v v

Therefore,

1 2

1 1

1

3 1

4

Iv v v

i i

i

R1 =3K R2 =1K

Iv t

2.0 V

i

Dv

0iDi

1v 2v

1i 2i

R1 =3K R2 =1K

Iv t

2.0 V

i

Dv

0iDi

1v 2v

1i 2i



And so:

10 25 Ii . v

meaning:

2 11 0 25 Iv . vi

Now you can write a second KVL:

2 22 0 2 2 0 25i i

D D Iv v v v . v

So you have concluded that the (reverse biased) ideal diode

voltage is: 2 0 25i

IDv . v

This result is true, provided that(!) your reverse bias

assumption is correct ( 0iDv ) ! You can quickly solve the

inequality to determine WHEN this is true:

2 0 25 0

0 25 2

8

8

I

I

I

I

. v

. v

v

v

R1 =3K R2 =1K

Iv t

2.0 V

i

Dv

0iDi

1v 2v

1i 2i



Thus:

2 0 25iIDv . v when 8Iv

The important (very important!) thing to note here is that the

ideal diode voltage is negative only when the input voltage is

significantly positive (i.e., 8Iv ).

Moreover, the larger (i.e., more positive) the input voltage

becomes, the more negative the ideal diode voltage becomes!

Thus, you come to the conclusion for this specific circuit (it

may not be true for other circuits!), that the minimum ideal

diode voltage ( i minDv ) occurs when the input voltage is at its

maximum (most positive). I.E.:

2 0 25i min maxIDv . v

Q: But how do I know what maxIv is? You said that max

Iv A ; so

then what is the “right equation” to “use” ?

A: Sigh; there is no equation. You know what the input

voltage is—it is given in the problem statement:

+6

+12

vI (t)

t -6

-12

+18



Note that the input voltage changes with respect to time.

At first, the input voltage is 6.0 V , later on it is at -6.0 V

and then after that -12.0 V. The important question is: when

is the input voltage at its maximum (i.e., most positive)?

Hopefully, by simply looking at

the input voltage signal (no

equations required!) it is evident

that 18.0 V is the most positive

the input voltage ever is!

Thus,

18 0maxIv .

Note that 18 0 8 0maxIv . . ; the ideal diode is definitely

reverse biased!

And so, during the time period when the input voltage is at

this maximum value, the ideal diode voltage will be at its

minimum (i.e., most negative):

2 0 25

2 0 25 18 0

2 5

i min maxIDv . v

. .

. V

Thus, the Peak Inverse Voltage (PIV) is:

2 2 55i minDvP . V.IV

No

Equations

Required



To confirm this, let’s plot the ideal diode voltage as a function

of time.

Q: Wait! We only know the ideal diode voltage when 8 0

Iv . V .

What is the ideal diode voltage when 8 0Iv . V ?

A: Remember, the ideal diode is reverse biased when

8 0I

v . V . Thus, when 8 0Iv . V the ideal diode is forward

biased.

And you know the voltage of a forward biased ideal diode!

0iDv if forward biased

Thus, the continuous function relating the ideal diode voltage

to the input voltage is:

2 0 25 8 0

0 8 0

I IiD I

I

. v when v . V

v f v

when v . V

vI

-0.25

8.0

iDv



Evaluating this with the input signal:

you see that the input is greater than 8.0 V only during the

period when it is equal to 18.0 V ! Thus, only during this time

period is the ideal diode voltage non-zero.

Specifically, you have found that when 18 0maxI Iv v . , the

ideal diode voltage is 2 5iDv . V .

Thus, you have confirmed that 2 5i minDv . V , and so:

2 2 55i minDvP . V.IV

+2.5

+5.0

iDv t

t

-5.0

-2.5

+6

+12

vI (t)

t -6

-12

+18



Q: Wasn’t this a bit of an academic exercise; I mean after all, an ideal diode doesn’t have a breakdown region!

A: True enough, this was an academic exercise. It did this to

simplify the analysis a bit.

However, if you are analyzing a circuit with a junction diode,

you still need to first find the minimum ideal diode voltage i minDv of the ideal diode in the CVD model.

Then, the minimum voltage of the junction diode is simply the

minimum voltage across the CVD model, which (of course!) is:

0 7min i minD Dv v .

Thus, the PIV is:

0 7min i minD DPIV v v .

Now this PIV value had better be smaller than the Zener

breakdown voltage of the junction diode, or else breakdown

will occur!

Documents

4.5 Rectifier Circuitsjstiles/312/handouts/section_3_5...3/1/2012 section_3_5_Rectifier_Circuits.doc 1/3 Jim Stiles The Univ. of Kansas Dept. of EECS 4.5 Rectifier Circuits Reading